670 kg h–1 of a slurry containing 120 kg solute and 50 kg solvent is to be extra

Question

670 kg h–1 of a slurry containing 120 kg solute and 50 kg solvent is to be extracted. The maximum permitted amount of solute in the final raffinate is 5 kg h–1. When a simple mixer-settling unit is used to separate extract and raffinate, the amount of solvent retained by the solid is 50 kg. Assuming perfect mixing and a constant ratio of solvent in extract and raffinate, determine the number of stages and the strength of the total extract for each of the following conditions: (i) Simple multiple contact is used for the extraction with a solvent addition of 100 kg h–1 per stage (ii) The same total amount of solvent as in (i) is provided in countercurrent operation (iii) Half the total amount of solvent in (i) is used in countercurrent operation. (b) Which of these methods would you recommend for the extraction? Give three reasons for your choice. (c) Suggest two alternative ways of carrying out the extraction and the circumstances under which they might be used.

Explanation / Answer

Since KD = y2/x2, y2=KDx2 Refer to EQ. 1
So, L(x1-x2)=V(KDx2)
x2[(VKD/L )+ 1)]= x1, where VKD/L = E Refer to EQ. 2
E= (6)(80)/100 = 4.8
x2/x1 = 1/ (1+E) Refer to EQ. 3 (frac. of penicillin in final raffinate phase from
raffinate phase in 1st stage) = frac. remaining)
= 1/ (1+ 4.8)
= 0.1724
x2/x0 = (x2/x1) * (x1/x0)
= (0.1724) * (0.1724)
= 0.0297 (frac. of penicillin in final raffinate phase from feed phase = frac.
remaining from )
Fraction of penicillin recovered = Fraction of penicillin in extract phase from
feed phase
= 1- 0.0297
= 0.9703
= 97.0%

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